Problem:
A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled A. The three vertices adjacent to vertex A are at heights 10,11, and 12 above the plane. The distance from vertex A to the plane can be expressed as tr−s​​, where r,s, and t are positive integers. Find r+s+t.
Solution:
Let the vertices of heights 10,11, and 12 above the plane be labeled B, C, and D, respectively. Note that these vertices are the vertices of an equilateral triangle with side length 102​. Set up a coordinate system where the given plane is the graph of z=−10,B is at (0,0,0),C is at (7,1023​​,1), and D is at (14,0,2). Suppose vertex A is at (x,y,z). Because A is a distance 10 from vertex B, it follows that x2+y2+z2=100. Because the vectors from B to A,C to A, and D to A are mutually perpendicular, their dot products are all zero, or (x,y,z)⋅(x−14,y,z−2)= 0 and (x,y,z)⋅(x−7,y−1023​​,z−1)=0. Thus 14x+2z=100 and 7x+1023​​y+z=100. These equations yield 7x+z=50 and y=532​​. Substituting x=750−z​ and y=532​​ into x2+y2+z2=100 and simplifying yields the quadratic equation 3z2−6z−95=0. This equation has solutions z=33±294​​. Because the plane is at z=−10 and vertex A lies below the xy plane, it follows that vertex A is at a height 10+33−294​​=333−294​​. The requested sum is therefore 33+294+3=330​.
Note that the vertex A is at a height of about 5.2845 above the plane.